Question

hw 2: problem 5
(1 point)
let $f(x)=3-x^2$. find each of the following:
(a) $\frac{f(8)-f(4)}{8-4} = $
(b) $\frac{f(4+h)-f(4)}{h} = $
(c) $\lim_{h\to0}\frac{f(4+h)-f(4)}{h} = $
note: you can earn partial credit on this problem.
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Explanation:

Step1: Calculate f(8) and f(4)

$f(8)=3-8^2=3-64=-61$
$f(4)=3-4^2=3-16=-13$

Step2: Solve part (A)

Substitute values into the expression:
$\frac{f(8)-f(4)}{8-4}=\frac{-61-(-13)}{4}=\frac{-48}{4}=-12$

Step3: Calculate f(4+h)

$f(4+h)=3-(4+h)^2=3-(16+8h+h^2)=-13-8h-h^2$

Step4: Solve part (B)

Substitute f(4+h) and f(4):
$\frac{f(4+h)-f(4)}{h}=\frac{(-13-8h-h^2)-(-13)}{h}=\frac{-8h-h^2}{h}=-8-h$ (where $h
eq0$)

Step5: Solve part (C)

Take the limit as $h\to0$:
$\lim_{h\to0}\frac{f(4+h)-f(4)}{h}=\lim_{h\to0}(-8-h)=-8$

Answer:

(A) $-12$
(B) $-8-h$
(C) $-8$